Quadratic Equations Class 10 Previous Year Questions (Chapter 4)

1. 1

   Step 1: Form the equations

   1 mark

   Let the greater number be x and the smaller number be y. Then x² − y² = 180 and y² = 8x.

   Why? A difference of squares is the larger square minus the smaller square, and '8 times' means multiplication by 8.

2. 2

   Step 2: Form the quadratic equation

   1 mark

   Substitute y² = 8x in x² − y² = 180: x² − 8x = 180, so x² − 8x − 180 = 0.

   Why? Substitution removes the variable y and leaves one quadratic equation in x, which can then be solved.

3. 3

   Step 3: Solve by factorisation

   1 mark

   x² − 8x − 180 = (x − 18)(x + 10) = 0, so x = 18 or x = −10.

   Why? A product of two factors is zero only when at least one factor is zero, so each linear factor gives one root.

4. 4

   Step 4: Select the positive value

   1 mark

   x = −10 is rejected because the numbers are positive. Hence x = 18, and y² = 8 × 18 = 144, so y = 12.

   Why? The word 'positive' in the question rejects the negative root; the relation y² = 8x then gives the smaller number.

5. 5

   Step 5: State the answer

   1 mark

   The two numbers are 18 and 12.

   Why? The final sentence states the two numbers that satisfy both given conditions.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Split as in the CBSE marking scheme, 2025 · Basic 430/1/2 Q35 (the marks split is taken from the 2025 Basic paper, where this question also appeared)

1. 1

   Step 1: Form the quadratic equation

   1 mark

   Let the two consecutive negative integers be x and x + 1. Then x² + (x + 1)² = 481.

   Why? Consecutive integers differ by 1, so if the smaller integer is x, the next is x + 1. The given sum of squares becomes the equation to solve.

2. 2

   Step 2: Simplify to standard form

   ½ mark

   x² + x² + 2x + 1 = 481 ⇒ 2x² + 2x − 480 = 0 ⇒ x² + x − 240 = 0.

   Why? Expanding (x + 1)² and collecting like terms gives a quadratic in standard form. Dividing by 2 keeps the coefficients smaller for factorisation.

3. 3

   Step 3: Factorise and find the roots

   1 mark

   x² + x − 240 = (x + 16)(x − 15) = 0, so x = −16 or x = 15.

   Why? Splitting the middle term factors the quadratic into two linear factors. A product is zero only when one factor is zero, so each linear factor gives one root.

4. 4

   Step 4: Choose the negative pair

   ½ mark

   Since the integers are negative, x = −16, so the integers are −16 and −15.

   Why? The word 'negative' rejects the positive root x = 15. The remaining root x = −16 gives the two consecutive negative integers −16 and −15.

Marks: 1 + ½ + 1 + ½ = 3 · Split as in the CBSE marking scheme, 2026 · Standard 30/3/2 Q26 (the marks split is taken from the 2026 Standard paper, where this question also appeared)

1. 1

   Step 1: Let the integers be x and x + 2

   1 mark

   Let the two consecutive even positive integers be x and x + 2.

   Why? A variable turns the given condition into an equation. Consecutive even integers differ by 2, so the next one is x + 2.

2. 2

   Step 2: Form the equation

   1 mark

   The sum of their squares is 884, so x² + (x + 2)² = 884.

   Why? Squaring each integer and adding them gives the sum of the squares. Equating this sum to 884 makes a quadratic equation in x.

3. 3

   Step 3: Simplify to standard form

   1½ marks

   x² + (x + 2)² = x² + x² + 4x + 4 = 884, so 2x² + 4x − 880 = 0. Dividing by 2 gives x² + 2x − 440 = 0.

   Why? Expanding (x + 2)² by (a + b)² = a² + 2ab + b² and collecting like terms gives the standard form ax² + bx + c = 0. Dividing every term by 2 keeps the roots the same and makes the factorisation easier.

4. 4

   Step 4: Solve by factorisation

   1 mark

   x² + 2x − 440 = (x + 22)(x − 20) = 0, so x = −22 or x = 20. Since x must be positive, x = −22 is rejected, so x = 20.

   Why? The expression is written as a product of two linear factors, which are factors of the form x − a. If a product is zero, one of its factors must be zero. The negative value does not fit the word positive, so it is rejected.

5. 5

   Step 5: State the integers

   ½ mark

   x = 20 and x + 2 = 22, so the two consecutive even positive integers are 20 and 22.

   Why? The smaller integer is 20, and adding 2 gives the next even integer 22. Together they satisfy the given condition.

Marks: 1 + 1 + 1½ + 1 + ½ = 5 · Split as in the CBSE marking scheme, 2024 Compartment · Basic 430/S/3 Q35

1. 1

   Step 1: Let the unknown

   1 mark (suggested)

   Let the number of articles produced in a day be x. Then the cost of production of each article = ₹(2x + 3).

   Why? The number of articles is the unknown, and the given relation lets the cost per article be written in terms of x.

2. 2

   Step 2: Form the quadratic equation

   1 mark (suggested)

   Total cost = x(2x + 3) = 90.

   Why? Total cost is the number of articles multiplied by the cost of one article.

3. 3

   Step 3: Write in standard form

   1 mark (suggested)

   2x² + 3x - 90 = 0.

   Why? Expanding and taking all terms to one side gives the standard form ax² + bx + c = 0, which is ready for factorisation.

4. 4

   Step 4: Factorise and solve

   1 mark (suggested)

   2x² + 3x - 90 = (2x + 15)(x - 6) = 0, so x = 6 or x = -15/2.

   Why? If the product of two factors is zero, one of the factors must be zero, so each linear factor gives a root.

5. 5

   Step 5: Reject the negative root and answer

   1 mark (suggested)

   The number of articles cannot be negative, so x = 6. Cost of each article = 2(6) + 3 = ₹15.

   Why? The negative root is impossible for a count of articles; substituting x = 6 in the cost expression gives the cost per article.

Marks: 1 + 1 + 1 + 1 + 1 = 5 · Suggested split (CBSE’s usual pattern for 5-mark questions)

1. 1

   Step 1: Form the equations

   1 mark (suggested)

   Let the sides of the two squares be x m and y m, with x > y. Then x² + y² = 468, and 4x − 4y = 24, so x − y = 6.

   Why? The area of a square is the square of its side, and its perimeter is four times its side, so the two given conditions become two equations in the unknowns.

2. 2

   Step 2: Solve by substitution

   1 mark (suggested)

   Putting x = y + 6 in x² + y² = 468: (y + 6)² + y² = 468, so 2y² + 12y − 432 = 0, i.e. y² + 6y − 216 = 0. Factorising: (y + 18)(y − 12) = 0. Hence y = −18 or y = 12.

   Why? Substitution removes one unknown; factorising the quadratic and equating each factor to zero gives its roots.

3. 3

   Step 3: Reject and conclude

   1 mark (suggested)

   The negative root is rejected, because a side cannot be negative. So y = 12, and x = y + 6 = 18. The lengths of the sides of the two squares are 18 m and 12 m.

   Why? Lengths must be positive, so the negative root cannot be a side; the equation x − y = 6 then gives the other side.

Marks: 1 + 1 + 1 = 3 · Suggested split (CBSE’s usual pattern for 3-mark questions)